Structural Stability and Load Analysis Project

1. WIND FORCES

2. wind load

4. Wind Load is an ‘Area Load’ (measured in PSF) which loads the surface area of a structure.

6. SEISMIC FORCES

7. seismic load

10. Seismic Load is generated by the inertia of the mass of the structure : VBASE Redistributed (based on relative height and weight) to each level as a ‘Point Load’ at the center of mass of each floor level: FX VBASE wx hx S(w h) VBASE = (Cs)(W) ( VBASE ) Fx =

11. Where are we going with all of this?

12. global stability & load flow (Project 1) tension, compression, continuity equilibrium: forces act on rigid bodies,and they move nearly imperceptibly boundary conditions: fixed, pin, or roller idealize member supports & connections external forces: areapplied to beams & columns as concentrated point loads & linear loads categories of external loading: DL, LL, W, E, S, H (fluid pressure) reactions: we use three equations of equilibrium to calculate these

13. internal forces:axial, shear, bending/flexure internal stresses:tension stress, compression stressshear stress, bending stress, stability, slenderness, and allowable compression stress member sizingfor flexure member sizing for combined flexure and axial stress (Proj. 2) Trusses (Proj. 3)

14. EXTERNAL FORCES

16. 200 lb ( + ) SM1 = 0 0= -200 lb(10 ft)+ RY2(15 ft) RY2(15 ft) = 2000 lb-ft RY2 = 133 lb ( +) SFY = 0 RY1 + RY2 - 200 lb = 0 RY1 + 133 lb- 200 lb = 0 RY1 = 67 lb ( +) SFX = 0 RX1 = 0 RX1 10 ft 5 ft RY2 RY1 200 lb 0 lb 10 ft 5 ft 67 lb 133 lb

19. w = 880lb/ft RX1 RY1 RY2 24 ft

20. resultant force - equivalent total load that is a result of a distributed line load resultant force = area loading diagram resultant force = w( L) RX1 RY1 RY2 24 ft

21. resultant force - equivalent total load that is a result of a distributed line load resultant force = area loading diagram resultant force = w( L) = 880 lb/ft(24ft) = 21,120 lb RX1 12 ft 12 ft RY1 RY2 24 ft

22. resultant force - equivalent total load that is a result of a distributed line load resultant force = area loading diagram resultant force = w( L) = 880 lb/ft(24ft) = 21,120 lb RX1 12 ft 12 ft RY1 RY2 24 ft ( + ) SM1 = 0

23. resultant force - equivalent total load that is a result of a distributed line load resultant force = area loading diagram resultant force = w( L) = 880 lb/ft(24ft) = 21,120 lb RX1 12 ft 12 ft RY1 RY2 24 ft ( + ) SM1 = 0 –21,120 lb(12 ft)+ RY2(24 ft) = 0

24. resultant force - equivalent total load that is a result of a distributed line load resultant force = area loading diagram resultant force = w( L) = 880 lb/ft(24ft) = 21,120 lb RX1 12 ft 12 ft RY1 RY2 24 ft ( + ) SM1 = 0 –21,120 lb(12 ft)+ RY2(24 ft) = 0 RY2(24 ft) = 253,440 lb-ft

25. resultant force - equivalent total load that is a result of a distributed line load resultant force = area loading diagram resultant force = w( L) = 880 lb/ft(24ft) = 21,120 lb RX1 12 ft 12 ft RY1 RY2 24 ft ( + ) SM1 = 0 –21,120 lb(12 ft)+ RY2(24 ft) = 0 RY2(24 ft) = 253,440 lb-ft RY2 = 10,560 lb

26. resultant force - equivalent total load that is a result of a distributed line load resultant force = area loading diagram resultant force = w( L) = 880 lb/ft(24ft) = 21,120 lb RX1 12 ft 12 ft RY1 RY2 24 ft ( + ) SM1 = 0 –21,120 lb(12 ft)+ RY2(24 ft) = 0 RY2(24 ft) = 253,440 lb-ft RY2 = 10,560 lb ( +) SFY = 0

27. resultant force - equivalent total load that is a result of a distributed line load resultant force = area loading diagram resultant force = w( L) = 880 lb/ft(24ft) = 21,120 lb RX1 12 ft 12 ft RY1 RY2 24 ft ( + ) SM1 = 0 –21,120 lb(12 ft)+ RY2(24 ft) = 0 RY2(24 ft) = 253,440 lb-ft RY2 = 10,560 lb ( +) SFY = 0 RY1 + RY2 – 21,120 lb = 0

28. resultant force - equivalent total load that is a result of a distributed line load resultant force = area loading diagram resultant force = w( L) = 880 lb/ft(24ft) = 21,120 lb RX1 12 ft 12 ft RY1 RY2 24 ft ( + ) SM1 = 0 –21,120 lb(12 ft)+ RY2(24 ft) = 0 RY2(24 ft) = 253,440 lb-ft RY2 = 10,560 lb ( +) SFY = 0 RY1 + RY2 – 21,120 lb = 0 RY1 + 10,560 lb– 21,120 lb = 0

29. resultant force - equivalent total load that is a result of a distributed line load resultant force = area loading diagram resultant force = w( L) = 880 lb/ft(24ft) = 21,120 lb RX1 12 ft 12 ft RY1 RY2 24 ft ( + ) SM1 = 0 –21,120 lb(12 ft)+ RY2(24 ft) = 0 RY2(24 ft) = 253,440 lb-ft RY2 = 10,560 lb ( +) SFY = 0 RY1 + RY2 – 21,120 lb = 0 RY1 + 10,560 lb– 21,120 lb = 0 RY1 = 10,560 lb

30. resultant force - equivalent total load that is a result of a distributed line load resultant force = area loading diagram resultant force = w( L) = 880 lb/ft(24ft) = 21,120 lb RX1 12 ft 12 ft RY1 RY2 24 ft ( + ) SM1 = 0 –21,120 lb(12 ft)+ RY2(24 ft) = 0 RY2(24 ft) = 253,440 lb-ft RY2 = 10,560 lb ( +) SFY = 0 RY1 + RY2 – 21,120 lb = 0 RY1 + 10,560 lb– 21,120 lb = 0 RY1 = 10,560 lb ( +) SFX = 0 RX1 = 0

31. resultant force - equivalent total load that is a result of a distributed line load resultant force = area loading diagram resultant force = w( L) = 880 lb/ft(24ft) = 21,120 lb RX1 12 ft 12 ft RY1 RY2 24 ft ( + ) SM1 = 0 –21,120 lb(12 ft)+ RY2(24 ft) = 0 RY2(24 ft) = 253,440 lb-ft RY2 = 10,560 lb ( +) SFY = 0 RY1 + RY2 – 21,120 lb = 0 RY1 + 10,560 lb– 21,120 lb = 0 RY1 = 10,560 lb ( +) SFX = 0 RX1 = 0 w = 880 lb/ft 0 lb 24 ft 10,560 lb 10,560 lb

32. SIGN CONVENTIONS (often confusing and frustrating) External –for solving reactions (Applied Loading & Support Reactions) + X pos. to right - X to left neg. + Y pos. up - Y down neg + Rotation pos. counter-clockwise - CW rot. neg. Internal –for P V M diagrams (Axial, Shear, and Moment inside members) Axial Tension (elongation) pos. | Axial Compression (shortening) neg. Shear Force (spin clockwise) pos. | Shear Force (spin CCW) neg. Bending Moment (smiling) pos. | Bending Moment (frowning) neg.

33. STRUCTURAL ANALYSIS: INTERNAL FORCES P V M

34. INTERNAL FORCES Axial (P) Shear (V) Moment (M)

35. + + + P M V

36. - - - P M V

37. RULES FOR CREATING P DIAGRAMS 1. concentrated axial load | reaction = jump in the axial diagram 2. value of distributed axial loading = slope of axial diagram 3. sum of distributed axial loading = change in axial diagram

38. - 0 + -10k -10k -10k -10k -20k -10k +20k +20k • 20k 0 • compression

39. RULES FOR CREATING V M DIAGRAMS (3/6) 1. a concentrated load | reaction = a jump in the shear diagram 2. the value of loading diagram = the slope of shear diagram 3. the area of loading diagram = the change in shear diagram

40. w = - 880 lb/ft 0 lb 24 ft 10,560 lb 10,560 lb P 0 0 Area of Loading Diagram -0.88k/ft * 24ft = -21.12k 10.56k + -21.12k = -10.56k +10.56k -880 plf = slope +10.56 k 0 V 0 +10.56 k -10.56k

41. RULES FOR CREATING V M DIAGRAMS, Cont. (6/6) 4. a concentrated moment = a jump in the moment diagram 5. the value of shear diagram = the slope of moment diagram 6. the area of shear diagram = the change in moment diagram

42. w = - 880 lb/ft 0 lb 24 ft 10,560 lb 10,560 lb P 0 0 0 0 Area of Loading Diagram -0.88k/ft * 24ft = -21.12k 10.56k + -21.12k = -10.56k +10.56k -880 plf = slope +10.56 k 0 V 0 +10.56 k zero slope -10.56k 63.36k’ -63.36 k-ft Slope initial = +10.56k +63.36 k-ft neg. slope pos. slope Area of Shear Diagram (10.56k )(12ft ) 0.5 = 63.36 k-ft M (-10.56k)(12ft)(0.5) = -63.36 k-ft

44. Wind Loading W2 = 30 PSF W1 = 20 PSF

45. Wind Load spans to each level 1/2 LOAD W2 = 30 PSF SPAN 10 ft 1/2 + 1/2 LOAD SPAN 10 ft W1 = 20 PSF 1/2 LOAD

46. Total Wind Load to roof level wroof= (30 PSF)(5 FT) = 150 PLF

47. Total Wind Load to second floor level wsecond= (30 PSF)(5 FT) + (20 PSF)(5 FT) = 250 PLF

48. wroof= 150 PLF wsecond= 250 PLF

49. seismic load